Question

If the discriminant b^2-4ac=0, the graph of f(x)=ax^2+bx+c, a≠0, will touch the x axis as it’s vertex

Answer 1

Answer:

The graph of $f(x)$ will touch the x axis at $\dfrac{-b}{2a}$ when $b^2-4ac=0$.

Step-by-step explanation:

The roots of quadratic equation $f(x)=ax^2+bx+c, a\neq 0$ is of the form,

$x=\dfrac{-b\pm \sqrt{b^2-4ac} }{2a}$

When  $b^2-4ac=0$, the above root become,

$x=\dfrac{-b}{2a}$

The roots of quadratic equation are nothing but the x-intercepts of the curve $f(x)$. So the curve $f(x)$ will touch the x-axis at $x=\dfrac{-b}{2a}$

Answer 2

The graph touches the x-axis at one point only.

• Consider a quadratic equation ax^2 + b x + c =0, the discriminant of a quadratic equation is given by the formula,

= > Discriminant = $\sqrt{b^2 - 4ac}$,

• If the value of the discriminant is greater than 0, the graph touches the x-axis at two points and the roots are real, distinct.
• If the value of the discriminant is equal to 0, the graph touches the x-axis only at one point, the roots are real and equal.
• If the value of the discriminant is less than 0, the graph doesn't intersect the x-axis at any point and the roots are imaginary.

• It is given that the discriminant value is 0, Hence, the graph of the given equation touches the x-axis only at one point and the equation has only a single root that repeats twice.

Therefore, the graph ax^2+b x+c = 0 touches the x-axis at one point and it touches the vertex.